Let S(k)=1+3+5+...+(2k-1)=3+k^{2}. Then which of the following is true?

  • Option 1)

    S(k)\Rightarrow S(k-1)\;

  • Option 2)

    \; \; S(k)\Rightarrow S(k+1)\; \; \;

  • Option 3)

    S(1)  is correct

  • Option 4)

    principle of mathematical induction can be used to prove the formula

 

Answers (1)
A Aadil Khan

As we learnt in 

Steps of Mathematical Induction (Verification step) -

Step 1: Verification step

Actual verification of the proposition of the starting value n=1

- wherein

2^{3n}-1 is divisible by 7

Put n=1, It Satisfies.

 

 and

Steps of Mathematical Induction (Induction Step) -

Step 2: Induction Step

Assuming the proposition to be true for n=k, and proving it is true for value     n=k+1

-

 

 and

Steps of Mathematical Induction (Generalization Step) -

Combine Verification step and Induction step

p(1) is true and p(n) is true for n+1 assuming it is true for n

-

 

 S (k) = 1+3+5+------+ (2k -1)

= 3 + k^{2}

S (1) = 1^{1} = 3 + 1 Which is not true

\because S (1) is not true

\therefore S (k) is true

1+ 3+5+ -------- + (2k - 1) = 3 +k^{2}

S (k+1) = 1+3+5+-----+(2k - 1)+(2k + 1)

=3+k ^{2} + 2k +1

= 3 + (k + 1)^{2}

S \left ( k+1 \right ) \Rightarrow S\left ( k \right )

 


Option 1)

S(k)\Rightarrow S(k-1)\;

This option is incorrect. 

Option 2)

\; \; S(k)\Rightarrow S(k+1)\; \; \;

This option is correct. 

Option 3)

S(1)  is correct

This option is incorrect. 

Option 4)

principle of mathematical induction can be used to prove the formula

This option is incorrect. 

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